Answer:
21
Step-by-step explanation:
Given,
Total number of marble except red and green  = 3 +3+1
                                        = 7
So, the total number of possible sets of five marbles such that none of them are green or red can be given by
[tex]n\ =\ ^7C_5[/tex]
  [tex]=\ \dfrac{7!}{(7-5)!.5!}[/tex]
  [tex]=\ \dfrac{7!}{5!.2!}[/tex]
 [tex]=\ \dfrac{42}{2}[/tex]
  = 21
So, the required number of possible sets are 21.